A218395 Numbers whose square is the sum of the squares of 11 consecutive integers.

11, 77, 143, 1529, 2849, 30503, 56837, 608531, 1133891, 12140117, 22620983, 242193809, 451285769, 4831736063, 9003094397, 96392527451, 179610602171, 1923018812957, 3583208949023, 38363983731689, 71484568378289, 765356655820823, 1426108158616757

Offset

0,1

Comments

a(n)^2 = Sum_{j=0..10} (x(n)+j)^2 = 11*(x(n)+5)^2 + 110 and b(n) = x(n)+5 give the Pell equation a(n)^2 - 11*b(n)^2 = 110 with the 2 fundamental solutions (11; 1) and (77; 23) and the solution (10; 3) for the unit form. A198949(n+1) = b(n); A106521(n) = x(n) and x(0) = -4.

General: If the sum of the squares of c neighboring numbers is a square with c = 3*k^2-1 and 1 <= k, then a(n)^2 = Sum_{j=0..c-1} (x(n)+j)^2 and b(n) = 2*x(n)+c-1 give the Pell equation a(n)^2 - c*(b(n)/2)^2 = binomial(c+1,3)/2. a(n) = 2*e1*a(n-k) - a(n-2*k); b(n) = 2*e1*b(n-k) - b(n-2*k); a(n) = e1*a(n-k) + c*e2*b(n-k); b(n) = e2*a(n-k) + e1*b(n-k) with the solution (e1; e2) for the unit form.

Links

Table of n, a(n) for n=0..22.

V. Pletser, Finding all squared integers expressible as the sum of consecutive squared integers using generalized Pell equation solutions with Chebyshev polynomials, arXiv preprint arXiv:1409.7972 [math.NT], 2014. See Table 1 p. 7.

Index entries for linear recurrences with constant coefficients, signature (0,20,0,-1).

Formula

a(n) = 20*a(n-2) - a(n-4); b(n) = 20*b(n-2) - b(n-4);

a(n) = 10*a(n-2) + 33*b(n-2); b(n) = 3*a(n-2) + 10*b(n-2).

a(n) = a(n-1) + 20*a(n-2) - 20*a(n-3) - a(n-4) + a(n-5).

G.f.: 11 * (1-x)*(1+8*x+x^2) / (1 - 20*x^2 + x^4).

With r=sqrt(11); s=10+3*r; t=10-3*r:

a(2*n) = ((11+r)*s^n + (11-r)*t^n)/2.

a(2*n+1) = ((77+23*r) * s^n + (77-23*r)*t^n)/2.

a(n) = 11 * A198947(n+1). - Bill McEachen, Dec 01 2022

Example

For n=6, Sum_{z=17132..17142} z^2 = 3230444569;
a(6) = sqrt(3230444569) = 56837;
b(6) = sqrt((a(6)^2-110)/11) = 17137; x(6) = b(6)-5 = 17132.

Maple

s:=0: n:=-1:
for j from -5 to 5 do s:=s+j^2: end do:
for z from -4 to 100000 do
  s:=s-(z-1)^2+(z+10)^2: r:=sqrt(s):
  if (r=floor(r)) then
    n:=n+1: a(n):=r: x(n):=z:
    b(n):=sqrt((s-110)/11):
    print(n, a(n), b(n), x(n)):
  end if:
end do:

Mathematica

LinearRecurrence[{0, 20, 0, -1}, {11, 77, 143, 1529}, 30] (* Harvey P. Dale, Aug 15 2022 *)

Crossrefs

c=2: A001653(n+1) = a(n); A002315(n) = b(n); A001652(n) = x(n).

Cf. A001032 (11 is a term of that sequence), A198947.

Sequence in context: A232032 A272395 A305727 * A208599 A325733 A059625

Adjacent sequences: A218392 A218393 A218394 * A218396 A218397 A218398

Keyword

nonn,easy

Author

Paul Weisenhorn, Oct 28 2012

Status

approved